【POJ3261】Milk Patterns【后缀数组】【二分】

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似乎是后缀数组的例题？

二分答案ans，将不小于ans的height分组，判断是否有一组个数大于k即可。

听说数字并不是非常大，所以直接把字符集大小设小了。当然也可以离散化。

#include <cstdio>#include <algorithm>using namespace std;const int maxn = 20005, maxm = 20005, M = 20000;int num[maxn], sa[maxn], rank[maxn], height[maxn];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}int wa[maxn], wb[maxn], wv[maxn], cnt[maxm];void SA(int *r, int n, int m) {int *x = wa, *y = wb;for(int i = 0; i < m; i++) cnt[i] = 0;for(int i = 0; i < n; i++) cnt[x[i] = r[i]]++;for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1];for(int i = n - 1; i >= 0; i--) sa[--cnt[x[i]]] = i;for(int j = 1; j < n; j <<= 1) {int p = 0;for(int i = n - j; i < n; i++) y[p++] = i;for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;for(int i = 0; i < n; i++) wv[i] = x[y[i]];for(int i = 0; i < m; i++) cnt[i] = 0;for(int i = 0; i < n; i++) cnt[wv[i]]++;for(int i = 1; i < m; i++) cnt[i] += cnt[i - 1];for(int i = n - 1; i >= 0; i--) sa[--cnt[wv[i]]] = y[i];swap(x, y);p = 1; x[sa[0]] = 0;for(int i = 1; i < n; i++)x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + j] == y[sa[i] + j] ? p - 1 : p++;if(p >= n) break;m = p;}}void calcHeight(int *r, int n) {int i, j, k;for(i = j = k = 0; i < n; height[rank[i++]] = k)for(k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++);}int n, k;bool check(int x) {int i = 1;while(1) {for(; i <= n && height[i] < x; i++);if(i > n) break;int cnt = 1;for(; i <= n && height[i] >= x; cnt++, i++);if(cnt >= k) return 1;}return 0;}int main() {n = iread(); k = iread();for(int i = 0; i < n; i++) num[i] = iread(); num[n] = 0;SA(num, n + 1, M);for(int i = 0; i <= n; i++) rank[sa[i]] = i;calcHeight(num, n);int l = 1, r = n;while(l <= r) {int mid = l + r >> 1;if(check(mid)) l = mid + 1;else r = mid - 1;}printf("%d\n", r);return 0;}

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